Question

Two copper circular discs are of the same thickness. The diameter of A is twice that of B. The moment of inertia of A as compared to that of B is

• A. Twice as large
• B. Four times as large
• C. 8 times as large
• D. 16 times as large

Answer 1

The correct option is D

16 times as large

Let ${}^{\prime }{\rho }^{\prime }$ be the density of copper circular discs.

In general, mass of disc of radius ${}^{\prime }{R}^{\prime }=M=\rho \left(\pi R^2\right)t$

$\therefore$ For a given disc $I=\frac{M{R}^2}{2}=\frac{\rho \pi t{R}^4}{2}$

$\therefore I\propto R^4$

$\Rightarrow \frac{I_A}{R_A^4}=\frac{I_B}{R_B^4}$

$\Rightarrow \frac{I_A}{I_B}=(\frac{R_A}{R_B}{)}^4=(\frac{2R_B}{R_B}{)}^4$

$\Rightarrow I_A=16I_B$